Bridges between different known integer sequences

Bridges between different known integer sequences

Roman Wituła, Damian Słota, Edyta Hetmaniok

Institute of Mathematics, Silesian University of Technology, Poland {roman.witula,damian.slota,edyta.hetmaniok}@polsl.pl

Abstract

In this paper a new method of generating identities for Fibonacci and Lu- cas numbers is presented. This method is based on some fundamental iden- tities for powers of the golden ratio and its conjugate. These identities give interesting connections between Fibonacci and Lucas numbers and Bernoulli numbers, Catalan numbers, binomial coefficients,δ-Fibonacci numbers, etc.

Keywords: Fibonacci and Lucas numbers, Bernoulli numbers, Bell numbers, Dobinski’s formula

MSC: 11B83, 11A07, 39A10

1. Introduction

The authors’ fascination with Fibonacci, Lucas and complex numbers has been reflected in the following two nice identities (discovered independently by Rabi- nowitz [10] and Wituła [7] and, probably, many other, former and future admirers of the Fibonacci and Lucas numbers):

(1+ξ+ξ⁴)ⁿ=Fn+1+Fn(ξ+ξ⁴) and (1+ξ²+ξ³)ⁿ =Fn+1+Fn(ξ²+ξ³), (1.1) whereξ⁵= 1,ξ∈Candξ6= 1, andFn denotes thenth Fibonacci number.

2. Basic identities

Let

α:= 2 cosπ

5 =1 +√ 5

2 and β:=−2 cos 2

5π

= 1−√ 5 2 .

Proceedings of the

15^thInternational Conference on Fibonacci Numbers and Their Applications Institute of Mathematics and Informatics, Eszterházy Károly College

Eger, Hungary, June 25–30, 2012

255

Then we have

α+β= 1, αβ=−1 (2.1)

Fn= αⁿ−βⁿ

α−β , n∈Z, (2.2)

Ln=αⁿ+βⁿ, n= 0,1,2, . . . , (2.3) whereLn denotes thenth Lucas number [3, 9].

Then, identities (1.1) can be written in the form

Fn+1+x⁻¹Fn=xⁿ, (2.4)

for everyx∈ {α, β}. In other words, we get the divisibility relation of polynomials (x²−x−1)|(xⁿ⁺¹−Fn+1x−Fn).

Similarly (by induction) we can generate the identity

Ln+1+x⁻¹Ln = (2x−1)xⁿ, (2.5) for everyx∈ {α, β}.This implies the following divisibility relation of polynomials

(x²−x−1)|((2x−1)xⁿ⁺¹−Ln+1x−Ln).

Remark 2.1. If the values Fn and Ln were defined for real subscripts n ∈ [0,1) (see [15]), then from formulae (2.4) and (2.5) we could easily extend these definitions for any other real subscripts.

In particular, if functions[0,1]3n7→Fn and[0,1]3n7→Ln are continuous, then from formulae (2.4) and (2.5) we could obtain the continuous extensions of these functions. With this problem also some special problem is connected (see Corollary 2.6 – Dobinski’s formula problem).

Immediately from identities (2.4) and (2.5) the next result follows.

Theorem 2.2 (Golden ratio power factorization theorem). Let {kn}^∞n=1 be a se- quence of positive integers. Then the following identities hold true

YN n=1

Fkn+1+

√5−1 2 Fkn

=1 +√ 5 2

^PN

n=1

kn

,

YN n=1

Fkn+1−

√5 + 1 2 Fkn

=1−√ 5 2

^PN

n=1

kn

,

or in equivalent compact form YN n=1

Fkn+1+x⁻¹Fkn

=x

PN n=1

kn

,

YN n=1

Lkn+ (2x−1)Fkn

= 2^Nx

PN n=1

kn

,

for everyx∈ {α, β},and YN

n=1

Lkn+1+

√5−1 2 Lkn

= √

5N1 +√ 5 2

^PN

n=1

kn

,

YN n=1

Lkn+1−

√5 + 1 2 Lkn

= −√

5N1−√ 5 2

^PN

n=1

kn

,

or in equivalent compact form YN n=1

Lkn+1+x⁻¹Lkn

= (2x−1)^Nx

PN n=1

kn

,

for everyx∈ {α, β}.The above identities are called "Golden Gate" relations.

We note that these identities act as links between Fibonacci and Lucas se- quences and many other special sequences of numbers, especially many known linear recurrence sequences. Now we will present the collection of such relations.

First let us consider the Bernoulli numbersBrdefined by the following recursion formula [6, 11]:

B0= 1, n

n−1

Bn−1+ n

n−2

Bn−2+. . .+ n

0

B0= 0, n= 2,3, . . . (we note that B2k+1 = 0, k = 1,2, . . .). Moreover, Bk(y) denotes here the k-th Bernoulli polynomial defined by

Bk(y) = Xk l=0

k l

Bly^k−l.

Corollary 2.3 (A bridge between Fibonacci, Lucas and Bernoulli numbers). We have

NY−1 n=1

F_n^k₊₁+x⁻¹F_n^k

=x

RN 0

Bk(y)dy

,

NY−1 n=1

L_n^k+ (2x−1)F_n^k

= 2^N−1x

NR

0

Bk(y)dy

and

NY−1 n=1

Ln^k+1+x⁻¹Ln^k

= (2x−1)^N−1x

NR

0

Bk(y)dy

, for everyx∈ {α, β}.

Proof. The identities result from the following known relation [6, 11]:

NX−1 n=1

n^k = ZN

0

Bk(y)dy= Xk r=0

k r

Br N^k−r+1 k−r+ 1.

Corollary 2.4(A bridge between Fibonacci numbers, Lucas numbers and binomial coefficients). We have

b(n+1)/2Y c k=1

F(^n−kk−1)⁺¹+x⁻¹F(^n−kk−1)

=x^Fn,

b(n+1)/2Y c k=1

L(ⁿk⁻−^k1)±√

5F(ⁿk⁻−^k1)

= 2^b(n+1)/2c1±√ 5 2

Fn

,

b(n+1)/2Y c k=1

L(ⁿk⁻−^k1)⁺¹+x⁻¹L(ⁿk⁻−^k1)

= (2x−1)^b(n+1)/2cx^Fn,

for everyx∈ {α, β}.

Proof. All the above identities follow from relation (see [9]):

Fn=

b(n+1)/2X c k=1

n−k k−1

.

Note that similar and simultaneously more general relations could be obtained for the incomplete Fibonacci and Lucasp−numbers (see [12, 13]).

Next corollary concerns the Catalan numbers defined in the following way Cn:= 1

n+ 1 2n

n

, n= 0,1, . . .

Corollary 2.5(A bridge between Fibonacci numbers, Lucas numbers and Catalan numbers). We have

YN n=0

F1+CN−nCn+x⁻¹FCN−nCn

=x^CN+1, (2.6)

YN n=0

LC_N−nCn+ (2x−1)FC_N−nCn

= 2^N+1x^CN+1 (2.7)

and YN

n=0

L1+CN−nCn+x⁻¹LCN−nCn

= (2x−1)^N+1x^CN+1, (2.8)

for everyx∈ {α, β}.

Moreover, ifpis prime andp≡3 (mod4),then we have

p

qx²F1+Cp−1 2

+x FCp−1 2

=x

2+C(p−1)/2

p , (2.9)

pr2

F₁₊¹

2C_p2−1 2

+x⁻¹F¹

2C_p2−1 2

F₁₊(^p−1p−1

2 )+x⁻¹F(^p−1p−1 2 )

=x

1 2C

(p2−1)/2+(^p−1p−1 2 )

p2 ,

(2.10) for everyx∈ {α, β}.

Proof. Identities (2.6)-(2.8) can be obtained from the recursive relation forCn

CN+1= XN n=0

CN−nCn, N= 0,1, . . .

Whereas relations (2.9) and (2.10) result from the fact that if p is prime and p≡3 (mod4),thenp|(2 +Cp−1

2 )and p²|

1 2Cp2−1

2

+ ^pp⁻−¹1 2

(see [1]).

Next conclusion is connected with the Bell numbersBn, n= 0,1, ...[6].

Corollary 2.6 (A bridge between Fibonacci numbers, Lucas numbers and Bell numbers). We have

YN n=0

F(^Nn)^Bn+1+x⁻¹F(^Nn)^Bn

=x^BN+1,

YN n=0

L(^Nn)^Bn+ (2x−1)F(^Nn)^Bn

= 2^N+1x^BN+1,

YN n=0

L(^Nn)^Bn+1+x⁻¹L(^Nn)^Bn

= (2x−1)^N+1x^BN+1,

for everyx∈ {α, β}.

Proof. All the above identities follow from the well known recursive relation B0:= 1,

BN+1= XN n=0

N n

Bn, N= 0,1, . . .

We note that for the Bell numbers the following interesting relation, called Dobinski’s formula [6], holds:

BN = 1 e

X∞ k=0

k^N

k!, N = 0,1,2, . . .

In connection with the above formula we formulate a certain problem which can be expressed in the following way. Is it possible to generalize the definition of Fibonacci numbersFn onto real indices (of Lucas numbersLn, respectively) such that the following equality will be fulfilled:

Y∞ k=0

F₁₊e−1kN

k! +x⁻¹Fe−1kN k!

=x^BN,

for everyx∈ {α, β} andN ∈N, or Y∞

k=0

L₁₊e−1kN

k! +x⁻¹Le−1kN k!

2x−1 =x^BN, for everyx∈ {α, β} andN ∈N, respectively?

Next corollary concerns the connection with the δ-Fibonacci numbers defined by relations (see [14]):

an(δ) = Xn k=0

n k

Fk−1(−δ)^k (2.11)

and

bn(δ) = Xn k=1

n k

(−1)^k−1Fkδ^k, (2.12) forδ∈C.

Corollary 2.7 (A bridge between Fibonacci, Lucas and δ-Fibonacci numbers).

For positive integersδ andn we get Yn

k=0

F₁₊(ⁿk)^Fk−1δ^k+x⁻¹F(ⁿk)^Fk−1δ^k

=x^an(−δ),

Yn k=1

F₁₊(ⁿk)^Fkδk+x⁻¹F(ⁿk)^Fkδk

=x^−bn(−δ),

Yn k=0

L(ⁿk)^Fk−1δ^k+ (2x−1)F(ⁿk)^Fk−1δ^k

= 2ⁿ⁺¹x^an(−δ),

Yn k=1

L(ⁿk)^Fkδk±√

5F(ⁿk)^Fkδk

= 2ⁿx^−bn(−δ),

Yn k=0

L₁₊(ⁿk)^Fk−1δ^k+x⁻¹L(ⁿk)^Fk−1δ^k

= (2x−1)ⁿ⁺¹x^an(−δ),

Yn k=1

L₁₊(ⁿk)^Fkδk+x⁻¹L(ⁿk)^Fkδk

= (2x−1)ⁿx^−bn(−δ),

etc., for every x∈ {α, β}. Moreover, we define hereFn+1=Fn+Fn−1,n∈Z.

Let us note that similar relations we have for the incompleteδ−Fibonacci num- bersan,r(δ)and bn,s(δ)where

an,r(δ) :=

Xr k=0

n k

Fk−1(−δ)^k, 0≤r≤n,

bn,s(δ) :=

Xs k=1

n k

(−1)^k−1Fkδ^k, 1≤s≤n.

Now we consider ther−generalized Fibonacci sequence{Gn}defined as follows

Gn =





0, if 0≤n < r−1,

1, if n=r−1,

G_n−1+G_n−2+. . .+G_n−r, if n≥r.

Corollary 2.8 (A bridge between Fibonacci, Lucas and classicr-Fibonacci num- bers). Let r∈N, r≥2. Then the following identities hold true [8]:

F1+2^r−1Gn−r+x⁻¹F2^r−1Gn−r

^r−1Y

k=1

F

1+(

rP−1 i=k

2ⁱ⁻¹)G_n−r−k+x⁻¹F

(

rP−1 i=k

2ⁱ⁻¹)G_n−r−k

=x^Gn, for everyn≥2r−1, and

hYⁿ

k=0

F_1+G²_k+x⁻¹F_G²_ki

×h^r−1Y

i=2 n−iY

k=0

F1+GkGk+i+x⁻¹FGkGk+i

i

=x^GnGn+1,

the special case of which is the following Lucas identity Yn

k=1

F_1+F²

k +x⁻¹F_F²

k

=x^FnFn+1,

for everyx∈ {α, β}.

Corollary 2.9. We have also (x∈ {α, β}):

FFn+1+1+x⁻¹FFn+1 FFn−1+1+x⁻¹FFn−1

=x^Ln, LFn+1±√

5FFn+1

LFn−1±√ 5FFn−1

= 41±√ 5 2

^Ln

, LFn+1+1+x⁻¹LFn+1 LFn−1+1+x⁻¹LFn−1

= 5x^Ln,

sinceFn+1+Fn−1=Ln,n∈N. Furthermore, we have FLn+1+1+x⁻¹FLn+1 FL_n−1+1+x⁻¹FL_n−1

=x^5Fn,

LLn+1±√ 5FLn+1

LLn−1±√ 5FLn−1

= 41±√ 5 2

^5Fn

, LLn+1+1+x⁻¹LLn+1 LLn−1+1+x⁻¹LLn−1+1

= 5x^5Fn,

sinceLn+1+Ln−1= 5Fn,n∈N.

Remark 2.10. Note that Theorem 2.2 is connected, in some way, with the following very important Zeckendorf’s theorem [6]:

For every number n∈N there exists exactly one increasing sequence2≤k1<

. . . < kr, wherer=r(n)∈N, such that ki+1−ki≥2 fori= 1,2, . . . , r−1, and n=Fk1+Fk2+. . .+Fkr.

For example, we have

1000 = 987 + 13 =F16+F7,

that is √

5F987±L987 √

5F13±L13

= 2L1000±2√

5F1000=

= L987±√ 5F987

L13±√ 5F13

= 41±√ 5 2

¹⁰⁰⁰ .

3. Final remark

Finally, we note that identities (2.4), considered at the beginning of this paper, were discussed by many authors. For example, S. Alikhani and Y. Peng [2] basing on (2.4) have proven thatαⁿ,for everyn∈N, cannot be a root of any chromatic polynomial. Furthermore, D. Gerdemann [5] has used the first of identities (2.4) for analyzing the, so called, Golden Ratio Division Algorithm. Consequently, he has discovered a semi-combinatorial proof of the following beautiful theorem.

Theorem 3.1. For nonconsecutive integers a1, . . . , ak, the following two state- ments are equivalent (for every m∈N):

m Fn =Fn+a1+Fn+a2+. . .+Fn+ak, m=α^a1+α^a2+. . .+α^ak.

Acknowledgements. The Authors are grateful to the valuable remarks of the Referee which gave the possibility to improve presentation of the paper.

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